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I encountered this question today:

$$\sum_{x=1}^{n}(20x -17) = 133234$$ Find the value of $n$

I tried to reformat so I can understand the question much better:$$\sum_{x=1}^{n}(20x -17) = (20-17)+(40-17)+(60-17)+\dots+(20(n)-17) = 133234$$ I noticed a pattern: $3 + 23 + 43 + 63 + 83+ \dots = 133234$

Then, I am stuck here. Can anyone continue or give me a hint to this problem. I am half-way done to find $n$.

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  • $\begingroup$ What contest is this from? $\endgroup$ – Joel Reyes Noche Feb 13 '17 at 9:47
  • $\begingroup$ @JoelReyesNoche I mentioned on my other question you asked. Anyway, what's the downvote for? $\endgroup$ – Adola Feb 18 '17 at 2:34
  • $\begingroup$ Math contests usually include in the question the year they are held. Your question has the numbers $20$ and $17$ in it, implying that it is a question from a math contest being held this year. You should indicate in your question what contest the question is from so that answerers can know if the contest is still on-going. Because if the contest is still on-going and you are asking about it, then you are cheating. The downvote is for not mentioning what contest your question is from even after it was asked for. Your refusal to answer implies that you don't want to get caught cheating. $\endgroup$ – Joel Reyes Noche Feb 18 '17 at 13:07
  • $\begingroup$ @JoelReyesNoche It's from my school, they decided to open up a math contest in their city. If I mentioned the name of the school, wouldn't you be able to know about my insecure(s): (location, education, full name). What I can tell is that the contest was very strict on not using any devices. I'll gratefully take the downvote, but that's all what you need to know $\endgroup$ – Adola Feb 19 '17 at 1:47
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Note $$\sum_{x=1}^{n}(20x -17) = \left(20 \sum _{x=1}^n x \right)-17n=10n(n+1)-17n= 133234$$ From the fact that we know $$\sum_{x=1}^{n}=\frac{n(n+1)}{2}$$ As seen here. So we have a quadratic for $n$. Simplifying, we have $$10n^2-7n=133234$$ However, this quadratic has no integer solutions, which can be verified through the quadratic formula and Wolframalpha. So I suspect a typo in your question.

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  • $\begingroup$ @Adola Anyway, as it stands there is no solution. $\endgroup$ – S.C.B. Feb 5 '17 at 10:54
  • $\begingroup$ @Adola None of them work. I'm not going to edit this in, though. $\endgroup$ – S.C.B. Feb 5 '17 at 11:00
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The number of terms $=n

The first term $=20\cdot1-17=3,$ the last term $20n-17$

So, the sum $$=\dfrac n2\left(3+20n-17\right)=10n^2-7n$$

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